Forcing

If is finite then there is finite such that
Whenever countable Transitive,
then there is a countable Countable Transitive Model where

Fake proof

Find in some , such that

Lets assume .
We find in some such that

and find some such that is bijective.

Find, outside of , some injective.
Clearly, .
Take to be the transitive closure of (which is countable)
and form a countable transitive model of containing .
Then in , we have
But we have no control over and and whether
So we can’t do it like this.

Theorem (Forcing)

IF has a Countable Transitive Model then …

Proof

Assume is consistent and to contradiction that is not.
So find finite such that is inconsistent.
Take as above.
Find a countable Transitive Model .
Now there is a countable Transitive Model such that

which is a contradiction!!